What is the integral from 1 to 7 of 1/(sqrt(2x+2)) ?

The integral from 1 to 7 of 1/(sqrt(2x+2)) is 2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 7 of 1/(sqrt(2x+2))

\int_{1}^{7} \frac{1}{2 x + 2} d x

2

Solution steps

\int_{1}^{7} \frac{1}{2 x + 2} d x

Apply u-substitution

: \int_{4}^{16} \frac{1}{2 u} d u

= \int_{4}^{16} \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{4}^{16} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{4}^{16}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{4}^{16}

Apply radical rule:

= \frac{1}{2} [2 u]_{4}^{16}

Compute the boundaries:

4

= \frac{1}{2} \cdot 4

Simplify

= 2

\frac{d}{d x} (tan^{2} (x^{3}))

x + 1 y = 3, x + 9 = y^{2}

\int \frac{x ^{3} + x ^{2}}{x ^{2} - 9 x + 20} d x

6 - x^{2}, x^{2} - 2, 0, 3

\int (\frac{π}{2} - 5 x^{- 0.5}) d x

What is the integral from 1 to 7 of 1/(sqrt(2x+2)) ?
The integral from 1 to 7 of 1/(sqrt(2x+2)) is 2